Consider a block matrix of the following form $$M=\left[\begin{matrix}f_1 & g \\ -g^T & f_2\end{matrix}\right]$$ $f_1, f_2$ are assumed symmetric, positive definite, invertible.
Now consider the factorisation $$M=\left[\begin{matrix}1 & X_1 \\ X_2 & 1\end{matrix}\right]^{-1}\left[\begin{matrix}1 & X_2^T \\ X_1^T & 1\end{matrix}\right]$$ or, equivalently $$\left[\begin{matrix}1 & X_1 \\ X_2 & 1\end{matrix}\right]\left[\begin{matrix}f_1 & g \\ -g^T & f_2\end{matrix}\right]=\left[\begin{matrix}1 & X_2^T \\ X_1^T & 1\end{matrix}\right]$$ $$\left[\begin{matrix}f_1-X_1 g^T & g+X_1 f_2 \\ X_2 f_1-g^T & X_2 g + f_2\end{matrix}\right]=\left[\begin{matrix}1 & X_2^T \\ X_1^T & 1\end{matrix}\right]$$ The question is this: what conditions do $f_1,f_2,g$ need to obey such that $X_1,X_2$ as defined by the equations above exist?
In the scalar case (where $f_1,f_2,g$ are $1\times 1$ matrices) the problem is trivial: even though there are 4 equations in 2 unknowns, a single condition suffices: $f_1 f_2=1-g^2$. It is the generalisation of this condition to the matrix case that I seek.
In the case where $f_1,f_2,g$ are larger than $1\times 1$, the problem becomes complicated. It is no longer easy to just solve two of the four equations for $X_1,X_2$, and check the solution obeys the other two. Solving for $X_1, X_2$ requires the solution of a Sylvester equation: $$\left\{ \begin{matrix}g+X_1 f_2=X_2^T \\ X_2 f_1-g^T=X_1^T\end{matrix} \right.$$ $$\left\{ \begin{matrix}X_1 =(X_2^T-g)f_2^{-1} \\ X_2 f_1-g^T=X_1^T\end{matrix} \right.$$ $$\left\{ \begin{matrix}X_1 =(X_2^T-g)f_2^{-1} \\ X_2 f_1-g^T=f_2^{-1} (X_2-g^T)\end{matrix} \right.$$ $$\left\{ \begin{matrix}X_1 =(X_2^T-g)f_2^{-1} & \\ X_2 f_1-f_2^{-1} X_2=-f_2^{-1} g^T+g^T & \text{Sylvester equation}\end{matrix} \right.$$ Though it might be possible to solve this and continue as in the scalar case, that seems both (1) extremely difficult and (2) overkill, given that I am not interested in the solutions themselves, but in whether or not they exist.
This question is a bit of a long shot, but any ideas or hints would be appreciated.